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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Example 2</dfn> A body of constant mass <span class="process-math">\(m\)</span> is projected vertically upward from the surface of the earth with an initial velocity <span class="process-math">\(v_0\text{.}\)</span> Neglecting the air resistance but taking into account the variation of the earth gravitational field with altitude. Finda) The expression for the velocity during the motion;b) The maximum height attained by the body;c) The smallest initial velocity for which the body will not return to earth.<dfn class="terminology">Solution:</dfn> From Newton’s second law, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation}
\mathrm{Gravity}=-\frac{mgR^2}{(R+x)^2}=m\frac{\textrm{d} v}{\textrm{d} t}.\tag{2.5.5}
\end{equation}
</div>
<p class="continuation">There are two unknown functions <span class="process-math">\(x(t)\)</span> and <span class="process-math">\(v(t)\)</span> in the above differential equation. And there exists a relation between these two unknown functions</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation}
\frac{\textrm{d} v}{\textrm{d} t}=\frac{\textrm{d} x}{\textrm{d} t} \frac{\textrm{d} v}{\textrm{d} x}=v \frac{\textrm{d} v}{\textrm{d} x}.\tag{2.5.6}
\end{equation}
</div>
<p class="continuation">Taking (<a href="" class="xref" data-knowl="./knowl/eq2_22.html" title="Equation 2.5.6">(2.5.6)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq2_21.html" title="Equation 2.5.5">(2.5.5)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation}
-\frac{gR^2}{(R+x)^2}=v\frac{\textrm{d} v}{\textrm{d} x}~\rightarrow~-\frac{gR^2 \textrm{d} x}{(R+x)^2}=v \textrm{d} v~\rightarrow~\frac{gR^2}{R+x}+C=\frac{v^2}{2}.\tag{2.5.7}
\end{equation}
</div>
<p class="continuation">Initially, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
x(0)=0, \quad v(0)=v_0.
\end{equation*}
</div>
<p class="continuation">This implies at <span class="process-math">\(x=0\text{,}\)</span> <span class="process-math">\(v=v_0\)</span> which further gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
\frac{gR^2}{R+0}+C=\frac{1}{2}v_0^2~\rightarrow~C=\frac{1}{2} v_0^2-gR.
\end{equation*}
</div>
<p class="continuation">From (<a href="" class="xref" data-knowl="./knowl/eq2_23.html" title="Equation 2.5.7">(2.5.7)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
\frac{1}{2} v^2=\frac{gR^2}{R+x}+\frac{1}{2} v_0^2-gR~\rightarrow~v=\pm \sqrt{\frac{2gR^2}{R+x}+v_0^2-2gR}.
\end{equation*}
</div>
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